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Newton's Method tahn(x)

  1. Feb 4, 2013 #1
    I was hoping someone could help me out with this problem. I need to find the values, that Newton's method converges for tanh(x).

    So far I set up the algorithm:

    x[itex]\tiny_{k+1}[/itex] = x[itex]_{k}[/itex] - [itex]\frac{tanhx}{sech^{2}x}[/itex]

    I simplified it to:

    x[itex]\tiny_{k+1}[/itex] = x[itex]_{k}[/itex] - ([itex]\frac{1}{2}[/itex]sinh(2x))

    And then I took the derivative to see when the absolute value of the derivative was less than one.

    |1 - cosh(2x)| < 1

    I did all of this and came up with the interval of convergence as being:

    [itex]\frac{1}{2}[/itex]ln(2[itex]\pm \sqrt{3}[/itex])

    These numbers are approximately .-658 and .658 respectively.

    However when I test out 1, it converges using Newton's method, yet it falls out of the range that I determined.

    Any help on the problem would be appreciated.
     
  2. jcsd
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